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i have plugged the x coordinates of the points in the x of the function and only graph that has the point (0,3) and (5,8) work and made both sides equal .. and that's answer choices A , B and , C so please help me.
http://assets.openstudy.com/updates/attachments/557a2a37e4b0636b8cc2f7da-doshka_syria-1434069633043-screenshot_1.png the second one here
@jim_thompson5910 @Luigi0210 @peachpi @ryamorgan284
Yes , i have uploaded them in order .. first one A and B and second one is C and D
How? -5 doesn't work if you plug it in the function it doesn't make both sides equal.
this question only concerns the x-values... \[x^2-4x+3=x+3\]
we know 0 would work because \[(0)^2-4(0)+3=0+3 \] \[3=3\] but since there are multiple coordinates, we need to figure out if there is more than one x value that satisfies the equation
there's 2 intersections... for your choices... so there's more than 1 common point
right, and i have done that and only the points (0,3) and (5,8) do satisfie that function
\[(5)^2-4(5)+3=5+3\] \[25-20+3=5+3\] \[5+3=5+3 \rightarrow 8=8\] yeah that's correct...
\[(-5)^2-4(-5)+3=-5+3\] \[25+20+3 \neq -2\] \[48 \neq -2\]
i think it's the second graph because if you only graph f(x) = x+3 it will only give you a linear graph that it's line only matches the second one and the last one but since the last one doesn't satisfie the function it's not correct and it's B
it is the second graph.. the coordinates of (0,3) and(5,8)
the mistake that i have made when i tried to graph it .. i graphed it like this f(x)=x2 − 4x + 3 = x + 3 .. i should have done the left side as one graph and the second one as another graph and see the final image.
Thanks a lot for your time and help. @UsukiDoll
oh... you have to split them up like f(x) = x+3 f(x) = x^2-4x+3 and compare the two together to see if there's a common point.
aha , and that's what i did and the end
it helps to make a table like x f(x) = x+3 f(x) =x^2-4x+3 0 3 3 1 4 0
that's how i knew B and those two points match and the others don't
whenever you see the same number... it's a solution to both f(x) =x+3 and f(x) = x^2-4x+3